Logarithms - Show that the expression is true

1. Dec 14, 2012

1. The problem statement, all variables and given/known data

Show that the following expression is true:

0.2*0.8 = -14dB + (-1.94dB)

2. Relevant equations

1 dB = 10log10(x) (where x is a ratio of two quantities)

10log10(ab) = 10log10(a) + 10log10(b)

3. The attempt at a solution

0.2*0.8 = -14dB + (-1.94dB)

0.16 = -15.94dB

___________________________

10log10(x)=-15.94

x=10-15.94/10=0.0255
___________________________

0.16 ≠ 0.0255 (so either my math is wrong or the expression is not true...)

2. Dec 14, 2012

Staff: Mentor

This doesn't make any sense. The left side is .16 with no units and the right side is -15.94, in units of dB.

Is this actually how the problem is presented?

3. Dec 14, 2012

Yes. I'm enrolled in an online college and just about every test, quiz, or homework I attempt, I find a plethora of inexplicable things like this. I'll go back to the professor to get an idea of what is missing, because something IS missing...

4. Dec 14, 2012

On your note about units... Decibels are unitless (dimensionless)... So aside from that, am I missing something else?

5. Dec 14, 2012

Staff: Mentor

I guess that the idea is to convert each dB value using the conversion formula, and see if the two dB values add up to .16.

6. Dec 14, 2012

skeptic2

What do you get for the logarithm of 0.8?
Now let's multiply each of those by 10, not 20.
and take the anti-log of the sum.

7. Dec 15, 2012

10log10(0.2) = -6.9897 dB
10log10(0.8) = -0.9691 dB

-6.9897 dB + -0.9691 dB = -7.9588 dB

10-7.9588/10=0.16

But that has not proven anything about the left side equaling the right side. It has merely manipulated the left side and we are back to the original question.

8. Dec 15, 2012

skeptic2

You're right. Where did the -14 dB and the -1.94 dB come from? If they are part of the problem then either they are not equal or you should have used 20*Log instead of 10*Log, but I see nothing in the problem to indicate that.

9. Dec 15, 2012

I asked the professor what was going on with the problem and he said that the math I did was right but the problem was not formed properly.

The thing with online schools is some of them have a group of staff members who create the homework/labs/tests and a separate group who instructs. This is an ABET-accredited school, which is good because I'm going to be an Electrical Engineer at the end of it all.

I see what you mean about 20*log(something). The "20" would indicate that two voltage levels were being compared because the formula for voltage gain has a "20" in it.